Question

我想使用POST方法向服务器发送参数。

我正在使用ServiceHandler.java文件来处理请求

public class ServiceHandler {

    static String response = null;
    public final static int GET = 1;
    public final static int POST = 2;

    public ServiceHandler() {

    }

    /**
     * Making service call
     * @url - url to make request
     * @method - http request method
     * */
    public String makeServiceCall(String url, int method) {
        return this.makeServiceCall(url, method, null);
    }

    /**
     * Making service call
     * @url - url to make request
     * @method - http request method
     * @params - http request params
     * */
    public String makeServiceCall(String url, int method,
                                  List<NameValuePair> params) {
        try {
            // http client
            DefaultHttpClient httpClient = new DefaultHttpClient();
            HttpEntity httpEntity = null;
            HttpResponse httpResponse = null;

            // Checking http request method type
            if (method == POST) {
                HttpPost httpPost = new HttpPost(url);

                // adding post params
                if (params != null) {
                    httpPost.setEntity(new UrlEncodedFormEntity(params));
                }

                httpResponse = httpClient.execute(httpPost);

            } else if (method == GET) {
                // appending params to url
                if (params != null) {
                    String paramString = URLEncodedUtils
                            .format(params, "utf-8");
                    url += "?" + paramString;
                }
                HttpGet httpGet = new HttpGet(url);

                httpResponse = httpClient.execute(httpGet);

            }
            httpEntity = httpResponse.getEntity();
            response = EntityUtils.toString(httpEntity);

        } catch (UnsupportedEncodingException e) {
            e.printStackTrace();
        } catch (ClientProtocolException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }

        return response;

    }
}

我正在从我的活动文件中传递参数

// BACKGROUND EXECUTION STARTS
        @Override
        protected Void doInBackground(Void... arg0) {


            // Creating service handler class instance
            ServiceHandler sh = new ServiceHandler();

           List<NameValuePair> params = new ArrayList<NameValuePair>();
            params.add(new BasicNameValuePair("pkey", "pvalue"));
            sh.makeServiceCall(url, ServiceHandler.POST, params);


            // Making a request to url and getting response
            String jsonStr = sh.makeServiceCall(url, ServiceHandler.POST);

            Log.d("Response: ", "> " + jsonStr);

我可以连接到我的服务器，但它没有从我的应用程序接收任何参数。

有人可以修改它并发送工作代码谢谢

Answer 1

尝试使用此代码：

ServiceHandler sh = new ServiceHandler();
List<NameValuePair> params = new ArrayList<NameValuePair>();

@Override
protected void onPreExecute() {
    // TODO Auto-generated method stub
    super.onPreExecute();


    params.add(new BasicNameValuePair("pkey", "pvalue"));

}

@Override
protected String doInBackground(String... param) {
        // TODO Auto-generated method stub
        String json;
        try {
            json = sh.makeServiceCall(url, ServiceHandler.POST, params);

            Log.d("Response: ", "> " + json);

        } catch (Exception e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }

        return null;
    }

@Override
protected void onPostExecute(String result) {
    // TODO Auto-generated method stub
    super.onPostExecute(result);


    }

Answer 2

你应该改为HttpUrlConnection：

public class MessageSender {
    private int responseCode;


    public boolean sendPost(YourParameter param) {

        HttpURLConnection connection;
        try {
            URL gcmAPI = new URL("your api url");
            connection = (HttpURLConnection) gcmAPI.openConnection();

            connection.setRequestMethod("POST");
            connection.setRequestProperty("Content-Type", "application/json");
                          connection.setDoOutput(true);

            ObjectMapper mapper = new ObjectMapper();
            mapper.setVisibility(PropertyAccessor.FIELD, JsonAutoDetect.Visibility.ANY);
            DataOutputStream dataOutputStream = new DataOutputStream(connection.getOutputStream());

            mapper.writeValue(dataOutputStream, param);

            dataOutputStream.flush();
            dataOutputStream.close();

            responseCode = connection.getResponseCode();
        } catch (IOException e) {
            e.printStackTrace();
        }
        if (responseCode == 200) {
            Log.i("Request Status", "This is success response status from server: " + responseCode);
            return true;
        } else {
            Log.i("Request Status", "This is failure response status from server: " + responseCode);
            return false;
        }
    }
}

在您的活动中：

MessageSender mgsSender = new MessageSender();
                    new AsyncTask<Void, Void, Void>() {
                        @Override
                        protected Void doInBackground(Void... params) {

                            mgsSender.sendPost(mgsContent);

                            }
                            return null;
                        }
                    }.execute();

Answer 3

/ *在服务处理程序类* /

上更改此行

public JSONObject makeHttpRequest(String url, String method,
List<NameValuePair> params) {
if(method == "POST"){
}else if(method == "GET"){
}
}

/ *在doinbackground中的ur活动类* /

sh.makeServiceCall(url,"POST", params);

在Android中使用POST方法发送参数

3 个答案: